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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.

$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.

The other two roots are found by division.

$$ \require{enclose} \begin{array}{rll} x^2 && \hbox{} \\[-3pt] x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt] \underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt] 2 +i.x^2 \end{array} $$

$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$ How to pursue by this or some other approach?

dxiv
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jiten
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7 Answers7

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As the coefficients are real thus the complex roots would appear in conjugate pair. Thus the second root would be $1-i$ and the third root (that is $-1$) can be calculated by using the fact that the sum of the roots of a cubic equation $ax^3+bx^2+cx+d$ is $-b/a$.

  • Long division wouldn't help here in this case? – jiten Jun 03 '22 at 02:47
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    You can certainly use long division, and if that's how you want to approach this problem, you can certainly do so as I did in my answer above. However, what RamanujanXXV suggests is also correct, as a special case of Vieta's Formula for $n=3$. This tells you what the roots should sum to, and since you know what 2 of the roots are, this leaves the final root as he describes! – JJ Hoo Jun 03 '22 at 02:48
  • @Jiten It would work as clarified by JJ Hoo. –  Jun 03 '22 at 02:55
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$x^3-x^2+2= (x-1-i)\,x^2 + 2 + i\,x^2\;$ How to pursue by this or some other approach?

The posted answers cover the other approaches.

OP's approach is also salvageable, but first the polynomial division must be taken to completion:

$$ x^3-x^2+2= \left(x-1-i\right)\left(x^2 + ix - 1 + i\right) $$

The quotient can be further factored either with the quadratic formula, or directly "by inspection":

$$ x^2 + ix - 1 + i = (x^2-1) + i(x+1)=(x+1)(x-1)+i(x+1)=(x+1)(x-1+i) $$

dxiv
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  • I was also thinking of completing the OP's method in my answer .Should I leave it now as it would be a copy of yours ? –  Jun 03 '22 at 03:05
  • @RamanujanXXV That's fine, if you add it to your answer I'll remove mine. – dxiv Jun 03 '22 at 03:06
  • No Sir, deleting your answer wouldn't be fair as you posted it first. I would better drop my idea.Thanks –  Jun 03 '22 at 03:10
  • @RamanujanXXV No problem either way. Just meant to answer the "by this approach" part of OP's question. – dxiv Jun 03 '22 at 03:13
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The equation $x^3 - x^2 + 2 = 0$ has three solutions. Therefore we can write it as $(x-a)(x-b)(x-c)=0$. Expanding the second expression and equating terms we get:

$a+b+c = 1$; $ab+bc+ac=0$; $abc=-2$

Now substitute $a = 1+i$. We get $b+c = -i$ and $bc=-1+i$. Which is easily solved.

M. Wind
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One thing to note is that complex roots of polynomials with real-valued coefficients ALWAYS come in pairs of complex conjugates (in other words, if $a+bi$ is a root, so is $a-bi$). In other words, since you know that $1+i$ is a root, this means you also know that $1-i$ is a root. If you are not aware of this fact, try to prove it for yourself or try to see why! If you would like, let me know and I can provide further justification.

Now that you know $1+i$ and $1-i$ are both roots, you can now see that $x-1-i$ and $x-1+i$ are both linear factors!

$$(x-1-i)(x-1+i)=x^2-2x+2$$

Now, doing the same long division you provide above, you should see that the final linear term is $x+1$!

JJ Hoo
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    Complex roots do NOT ALWAYS come in pairs. It's true only when the coefficients are reals. As an example, consider $x^2-ix=0$. – ultralegend5385 Jun 03 '22 at 02:49
  • Apologies, I meant to say that they always come in pairs when the polynomial lives in $\mathbb{R}[x]$, as this is what the OP had. I'll edit my post to reflect this! – JJ Hoo Jun 03 '22 at 02:50
  • Confused as my polynomial is different than $x^2-2x+2$. Also, if have just found a new polynomial (factor of $x^3-x^2+2$) using product of two conjugate roots, then what is quotient given on division of $x^3-x^2+2$ by $x^2-2x+2$. – jiten Jun 03 '22 at 02:51
  • Could you clarify what you mean, Jiten? – JJ Hoo Jun 03 '22 at 02:54
  • $x^2-2x+2$ is simply from the computation I show above. Perhaps this might clarify things.

    We know that your original polynomial can be written as the product of 3 linear terms, $(x-a)(x-b)(x-c)$, where $a,b,c$ are all in the complex numbers. Since we know that $a$ is $1-i$, we can say that $b$ is $1+i$ by the logic above, since $1+i$ is the complex conjugate of $1-i$. This means that we can solve for $(x-c)$ by dividing your original polynomial by $(x-a)(x-b)$ by long division. The polynomial i referenced, $x^2-2x+2$, is exactly $(x-a)(x-b)$.

     – JJ Hoo Jun 03 '22 at 02:56
  • Yes, the third root is real $x=-1$. But, is directly doing long division with one imaginary root not a valid idea then? – jiten Jun 03 '22 at 03:01
  • Exactly! So that tells us that our roots are $1+i$, $1-i$, and $-1$. – JJ Hoo Jun 03 '22 at 03:02
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    Directly doing long division is valid, but the issue is that you are not able to easily extract another linear term out to find a root by factoring. You can, however, use the quadratic formula as a next step from what you have! – JJ Hoo Jun 03 '22 at 03:05
  • Kindly elaborate the use of quadratic formula as a possible option. – jiten Jun 03 '22 at 03:07
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    @jiten Problem is that what you posted is not the long division. When you divide by a linear factor, the remainder should be a constant (and in this case you know it should be $0$). But yours looks to be $2+ix^2$, except that's not a constant and not the remainder. – dxiv Jun 03 '22 at 03:09
  • @dxiv so you mean the divisor shouldn't contain non linear terms. Then imaginary terms are not allowed? – jiten Jun 03 '22 at 03:12
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    @jiten Doesn't matter whether you work in reals or complexes, but when you divide a cubic by a linear the quotient must be a quadratic and the remainder must be a constant. The quotient and remainder will have the same kind of coefficients as the divisor and dividend, whether real or complex. – dxiv Jun 03 '22 at 03:15
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    As further clarification to what dxiv has mentioned, you did not perform your long division to completion. you stopped with $2+ix^2$, but you can further continue this by multiplying by ix, as is typical in long division. – JJ Hoo Jun 03 '22 at 03:18
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    Using an example with integers, suppose you were computing $\frac{214}{5}$ using long division. Your first step would be to place a 4, giving you $214-200=14$. You would not simply write your solution as $40+\frac{14}{5}$. While this is not an incorrect answer, you would, with long division, proceed one step further to get $40+2+\frac{4}{5}=42+\frac{4}{5}$. Moving to long division in polynomials does not change this process. Since you are dividing by a linear term (linear meaning that it is of the form $(x-a)$ for some $a\in\mathbb{C}$, this means that your remainder MUST be one degree smaller – JJ Hoo Jun 03 '22 at 03:21
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One of the other roots is $1-i,$ since complex roots of real polynomials come in complex conjugate pairs.

The sum of the roots is $-1$ times the coefficient of $x^2.$ So, if $r$ is the third root, $(1+i)+(1-i)+r=1,$ or $r=-1.$

Thomas Andrews
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Maybe all has been said already.

Polynomial with real coefficients:

1)Complex roots occur in pairs, one is the complex conjugate of the other. (Complex root theorem)

$x_1=1+i,$ $x_2=1-i;$

2)By inspection the real root is $x_3=-1.$

Peter Szilas
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We'll pretend that we don't already know (by a familiar theorem) that $ \ \overline{1 + i} \ $ is also a zero of $ \ x^3 - x^2 + 2 \ $ and apply a method you looked at in this problem.

We will re-arrange the polynomial equation $ \ x^3 - x^2 + 2 \ = \ 0 \ $ as $ \ x^3 - x^2 \ = \ -2 $ $ \rightarrow \ x^2·(x - 1) \ = \ -2 \ \ . $ Since $ \ (1 + i)^2 \ = \ 1 - 1 + 2i \ \ , \ $ the factors in the equation for $ \ z_1 \ = \ 1 + i \ $ become $ \ (2i)·( \ [1 + i] - 1 \ ) \ = \ (2i)·(i) \ = \ -2 \ \ . \ $ But we can readily change the sign of both factors without altering the correctness of the equation, thus $ \ (-2i)·(-i) \ = \ -2 \ \ ; \ $ the second zero would be $ \ z_2 - 1 \ = \ -i \ \Rightarrow \ z_2 \ = \ 1 - i \ \ $ (and indeed, $ \ z_2^2 \ = \ (1 - i)^2 \ = \ 1 - 1 - 2i \ \ ) \ . $ As for the third zero, it is clear "by inspection" that $ (-1)^3 - (-1)^2 \ = \ (-1) - 1 \ = \ -2 \ \ . \ $ (We can confirm this since we have now established that $ \ (z - 1 - i)·(z - 1 + i) \ = \ z^2 - 2z + 2 \ $ is a factor of $ \ z^3 - z^2 + 2 \ \ , \ $ so polynomial division shows the third factor to be $ \ z + 1 \ \ . \ ) $